Mean temperature of winding when current falls

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DevonZA
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Homework Statement


Multiple-choice physics question: a 250V potential difference across a field winding at 15°C, current 5A falling to 3.91A, alpha15 = 1/254.5 per °C; options 18.45°C, 84.25°C, 100°C, 75°C.


Homework Equations


R2=R1(1+alpha(t2-t1))

The Attempt at a Solution


R1=250/5=50ohms
R2=250/3.91=63.94ohms

R2=R1(1+alpha15degrees(t2-t1))
63.94=50(1+1/254.5(t2-15))
t2=

Now I found this online but the answers provided still don't match, 84.25 being the closest.

upload_2016-5-1_12-12-12.png


When I manipulate the equation I get a totally different answer:
63.94=50(1+1/254.5(t2-15))
63.94/50=(1+1/254.5(t2-15))
1.2788/1+1/254.5=t2-15
1.274=t2-15
t2=1.274+15=16.274 degrees

Please help me understand where I am going wrong.
 
DevonZA said:
where I am going wrong.
DevonZA said:
63.94/50=(1+1/254.5(t2-15))
1.2788/1+1/254.5=t2-15
Worked solution from a textbook: Example 1.13 on a field winding's resistance at 15°C and an unknown mean temperature t2, with R1 = 250/5 = 50 Ω, R2 = 250/3.91 = 63.94 Ω, and 63.94 = 50[1 + (1/254.5)(t2 − 15)], giving t2 = 86°C.

Is this from B.L. Theraja?
 
DevonZA said:
Not sure I found it online
It is from B.L. Theraja. The page looks familiar. The red line in the above post is your error.
 
What have I done wrong though? Every way I enter it into my calculator yields an incorrect answer
 
DevonZA said:
63.94/50=(1+1/254.5(t2-15))
1.2788/1+1/254.5=t2-15[/\QUOTE]
This step is wrong. Check the sequence of operations.